# What are all the possible rational zeros for f(x)=4x^4+19x^2-63 and how do you find all zeros?

Aug 26, 2016

$x = \pm \frac{3}{2}$ or $x = \pm \sqrt{7} i$

#### Explanation:

$f \left(x\right) = 4 {x}^{4} + 19 {x}^{2} - 63$

By the rational roots theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 63$ and $q$ a divisor of the coefficient $4$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{4} , \pm \frac{1}{2} , \pm \frac{3}{4} , \pm 1 , \pm \frac{3}{2} , \pm \frac{7}{4} , \pm \frac{9}{4} , \pm 3 , \pm \frac{7}{2} , \pm \frac{9}{2} , \pm \frac{21}{4} , \pm 7 , \pm 9 , \pm \frac{21}{2} , \pm \frac{63}{4} , \pm 21 , \pm \frac{63}{2} , \pm 63$

We could try each of these in turn, but there are easier ways to find the zeros of $f \left(x\right)$.

Note that $f \left(x\right)$ is quadratic in ${x}^{2}$, so we can use the quadratic formula to find:

${x}^{2} = \frac{- 19 \pm \sqrt{{19}^{2} - 4 \left(4\right) \left(- 63\right)}}{2 \cdot 4}$

$= \frac{- 19 \pm \sqrt{361 + 1008}}{8}$

$= \frac{- 19 \pm \sqrt{1369}}{8}$

$= \frac{- 19 \pm 37}{8}$

i.e. ${x}^{2} = \frac{9}{4}$ or ${x}^{2} = - 7$

Since the result is rational, we could have found this using an AC method instead, but at least the quadratic formula gives it to us directly.

Hence $x = \pm \frac{3}{2}$ or $x = \pm \sqrt{7} i$